# Derive the molecular formula of a hydrocarbon, the mass fraction of hydrogen

**Derive the molecular formula of a hydrocarbon, the mass fraction of hydrogen in which is 14.3%, and the vapor density of this substance by air is 3.86.**

Given:

w% (H) = 14.3%

D (air) = 3.86

Find:

Formula -?

Solution:

1) Find the molecular weight of a hydrocarbon by its density in air:

D (air) = Mr (HC): Mr (air), – since air does not have an exact molecular weight, it is conventionally taken as 29:

Mr (HC) = D (air) * 29 = 3.86 * 29 = 112.

2) Knowing the mass fraction of hydrogen in the composition, we find its molecular weight and the number of atoms:

112 – 100%

x – 14.3%,

then x = 14.3% * 112: 100% = 16.

N (H) = 16: Mr (H) = 16: 1 = 16 atoms.

3) Then the molecular weight of carbon accounts for 112 – 16 = 96

N (C) = 96: Mr (C) = 96: 12 = 8 atoms.

We got a substance with the formula C8H16 – octene.

Answer: С8Н16 – octene.